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Chapter 12: Q. 53 (page 817)

The sum of the first 120 terms of the sequence 14, 16, 18, 20...…

Short Answer

Expert verified

The sum of the first 120 terms of the Arithmetic sequence is 15960

Step by step solution

01

Given information

Arithmetic sequence 14, 16, 18, 20...…

we need to find sum of first 120 terms

02

Sum

The formula to find out sum is

$S_{n} = \frac{n}{2} (2 a_{1} + (n - 1) d) w h e r e a_{1} i s t h e f i r s t t e r m d i s t h e d i f f e r e n c e b e t w e e n t h e t e r m s n i s t h e n u m b e r o f t e r m s$

03

common difference

given arithmetic sequence is 14,16,18, 20,....

common difference d= 16-14=2

first term is 14.

we need to find out sum of 120 terms, so n=12

04

Sum

Substitute all the values and find out sum

$S_{n} = \frac{n}{2} (2 a_{1} + (n - 1) d) S_{120} = \frac{120}{2} (2 (14) + (120 - 1) 2) S_{120} = 60 (28 + 238) S_{120} = 15960 S u m o f f i r s t 120 t e r m s = 15960$

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Most popular questions from this chapter

In Problems 51–66, determine whether each infinite geometric series converges or diverges. If it converges, find its sum.

$6 + 2 + \frac{2}{3} + . . . . .$

For the function $f (x) = \frac{x - 1}{x},$ find $f (2) a n d f (3)$ .

Find x so that x, x + 2, and x + 3 are consecutive terms of a geometric sequence.

In Problems 51–60, write out each sum.

$\sum_{k = 1}^{n} (k + 2)$

In Problems 71-82, find the sum of each sequence.

${\sum (5}_{k = 1}^{20} k + 3)$

See all solutions

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